Q. 37 L5.0( 1 Vote )

# Solve each of the following initial value problems: , y = 2 when  , y = 2 when Given and  This is a first order linear differential equation of the form Here, P = –3 cot x and Q = sin 2x

The integrating factor (I.F) of this differential equation is,   We have   [ m log a = log am]  I.F = cosec3x [ elog x = x]

Hence, the solution of the differential equation is,      Recall ycosec3x = 2(–cosec x) + c

ycosec3x = –2cosec x + c  y = –2sin2x + csin3x

However, when , we have y = 2 2 = –2(1)2 + c(1)3

2 = –2 + c

c = 4

By substituting the value of c in the equation for y, we get

y = –2sin2x + (4)sin3x

y = –2sin2x + 4sin3x

Thus, the solution of the given initial value problem is y = –2sin2x + 4sin3x

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Interactive Quiz on DIfferential CalculusFREE Class  Functional Equations - JEE with ease48 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 